For a velocity field
$$ \textbf G(x, y) = (3x^2 − 6y^2 + 1)\textbf i + (x + 4y − 12xy)\textbf j $$ show that the work done in moving a particle on the unit circle centred at (1, 0) taking an anticlockwise direction, starting and ending at the rightmost point of the circle, can be expressed as
$$W=\oint_{c} dx +\oint_{c} x dy $$
Thereafter compute W by evaluating the line integral.
I understand that the integral form is some type of $$\oint_{c} \textbf G \cdot d\textbf r$$ where $$ \textbf r(t) = (1+\cos t)\textbf i + \sin t \textbf j$$ this being my parametrization for the above mentioned circle, but I get lost from here.
1 Answer
You have at least two options: $$ \oint_C \vec{G}\cdot d\vec{r} = \int_0^{2\pi} \vec{G}(\vec{r}(t))\cdot \vec{r}'(t) dt $$ but you end up with a long integral.
A better option is to use the Green theorem:
$$ \oint_C \vec{G}\cdot d\vec{r} = \iint_D Q_x-P_y \;dA = \iint_D 1 -12y + 12 y \; dA = A(D) = \pi $$
